B E 2100 Lecture Notes - Lecture 6: Independent And Identically Distributed Random Variables, Western Uttar Pradesh, Bernoulli Distribution
21 Jun 2018
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Statistics and Their Distributions
The sample mean, regarded as a statistic (before a sample has been selected or an experiment carried
out), is denoted by ; the calculated value of this statistic is .
Similarly, S represents the sample standard deviation thought of as a statistic, and its computed value is
s.
If samples of two different types of bricks are selected and the individual compressive strengths are
denoted by X1, . . . , Xm and Y1, . . . , Yn, respectively, then the statistic
, the difference between the two sample mean compressive strengths, is often of great interest.
Statistics and Their Distributions
Any statistic, being a random variable, has a probability distribution. In particular, the sample mean
has a probability distribution.
Suppose, for example, that n = 2 components are randomly selected and the number of breakdowns
while under warranty is determined for each one.
Possible values for the sample mean number of breakdowns are 0 (if X1 = X2 = 0), .5 (if either X1 = 0 and
X2 = 1 or X1 = 1 and X2 = 0), 1, 1.5, . . ..
Statistics and Their Distributions
The probability distribution of specifies P( = 0),
P( = .5), and so on, from which other probabilities such as P(1 3) and P( 2.5) can be calculated.
Similarly, if for a sample of size n = 2, the only possible values of the sample variance are 0, 12.5, and 50
(which is the case if X1 and X2 can each take on only the values 40, 45, or 50), then the probability
distribution of S2 gives P(S2 = 0), P(S2 = 12.5), and P(S2 = 50).
Statistics and Their Distributions
The probability distribution of a statistic is sometimes referred to as its sampling distribution to
emphasize that it describes how the statistic varies in value across all samples that might be selected.
Random Samples
The probability distribution of any particular statistic depends not only on the population distribution
(normal, uniform, etc.) and the sample size n but also on the method of sampling.
Consider selecting a sample of size n = 2 from a population consisting of just the three values 1, 5, and
10, and suppose that the statistic of interest is the sample variance. If sampling is done “with
replacement,” then S2 = 0 will result if X1 = X2.
Random Samples
Random Samples
Conditions 1 and 2 can be paraphrased by saying that the Xi’s are independent and identically distributed
(iid).
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If sampling is either with replacement or from an infinite (conceptual) population, Conditions 1 and 2
are satisfied exactly.
These conditions will be approximately satisfied if sampling is without replacement, yet the sample size
n is much smaller than the population size N.
Example 5.21
A certain brand of MP3 player comes in three configurations: a model with 2 GB of memory, costing
$80, a 4 GB model priced at $100, and an 8 GB version with a price tag of $120.
If 20% of all purchasers choose the 2 GB model, 30% choose the 4 GB model, and 50% choose the 8 GB
model, then the probability distribution of the cost X of a single randomly selected MP3 player purchase
is given by
with = 106, 2 = 244
Example 5.21
Suppose on a particular day only two MP3 players are sold. Let X1 = the revenue from the first sale;
X2 = the revenue from the second.
Suppose that X1 and X2 are independent, each with the probability distribution shown in (2) [so that X1
and X2 constitute a random sample from the distribution (2)].
Example 5.21
Table below lists possible (x1, x2) pairs, the probability of each [computed using (2) and the assumption
of independence], and the resulting and s2 values
Example 5.21
Now to obtain the probability distribution of , the sample average revenue per sale, we must consider
each possible value and compute its probability. For example, = 100 occurs three times in the table
with probabilities .10, .09, and .10, so
Px (100) = P( = 100) = .10 + .09 + .10 = .29
Similarly,
pS2(800) = P(S2 = 800) = P(X1 = 80, X2 = 120 or X1 = 120, X2 = 80)
= .10 + .10 = .20
Example 5.21
The complete sampling distributions of and S2 appear in (3) and (4)
Example 5.21
From (3),
= (80)(.04) + . . . + (120)(.25) = 106 =
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Second, it appears that the distribution has smaller spread (variability) than the original distribution,
since probability mass has moved in toward the mean. Again from (3),
= (802)(.04) + + (1202)(.25) – (106)2
Example 5.21
The variance of is precisely half that of the original variance (because n = 2). Using (4), the mean value
of
S2 is
S2 = E(S2) = S2 pS
2(s2)
= (0)(.38) + (200)(.42) + (800)(.20) + 244 = 2
That is, the sampling distribution is centered at the population mean , and the S2 sampling
distribution is centered at the population variance 2.
Example 5.21
If there had been four purchases on the day of interest, the sample average revenue would be based
on a random sample of four Xi’s, each having the distribution (2).
More calculation eventually yields the pmf of for n = 4 as
Simulation Experiments
The second method of obtaining information about a statistic’s sampling distribution is to perform a
simulation experiment.
This method is usually used when a derivation via probability rules is too difficult or complicated to be
carried out. Such an experiment is virtually always done with the aid of a computer.
The Distribution of the Sample Mean
The Distribution of the Sample Mean
The standard deviation is often called the standard error of the mean; it describes the
magnitude of a typical or representative deviation of the sample mean from the population mean.
Example 5.25
In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic
emission (used to indicate crack initiation) is = 28,000, and the standard deviation of the number of
cycles is = 5000.
Let X1, X2, . . . , X25 be a random sample of size 25, where each Xi is the number of cycles on a different
randomly selected specimen.
Then the expected value of the sample mean number of cycles until first emission is E( ) = 28,000, and
the expected total number of cycles for the 25 specimens is
E(To) = n = 25(28,000) = 700,000.
Example 5.25
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Document Summary
The sample mean, regarded as a statistic (before a sample has been selected or an experiment carried out), is denoted by ; the calculated value of this statistic is . Similarly, s represents the sample standard deviation thought of as a statistic, and its computed value is. If samples of two different types of bricks are selected and the individual compressive strengths are denoted by x1, . , the difference between the two sample mean compressive strengths, is often of great interest. Any statistic, being a random variable, has a probability distribution. In particular, the sample mean has a probability distribution. Suppose, for example, that n = 2 components are randomly selected and the number of breakdowns while under warranty is determined for each one. Possible values for the sample mean number of breakdowns are 0 (if x1 = x2 = 0), . 5 (if either x1 = 0 and.
